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JEE Main20246 Apr 2024Evening ShiftPhysicsCurrent ElectricityActual

The number of electrons flowing per second in the filament of a 110 ~W bulb operating at 220 ~V is : ( . Given . e =1.6 10⁻¹⁹ C )

Options

  1. A6.25 10¹⁷
  2. B1.25 10¹⁹
  3. C6.25 10¹⁸
  4. D31.25 10¹⁷

Correct answer

D. 31.25 10¹⁷

Step-by-step solution

Power ( P )= V . I aligned & 110=(220)( I ) & I =0.5 ~A aligned Now, I= n e t aligned & 0.5= ( n t ) (1.6 10⁻¹⁹ ) & n t = 0.5 1.6 10⁻¹⁹ & n t =31.25 10¹⁷ aligned

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